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Question Number 88067 by A8;15: last updated on 08/Apr/20

Answered by MJS last updated on 08/Apr/20

x=±(√(a^2 +b^2 +c^2 ))  it is easy to see

$${x}=\pm\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see} \\ $$

Commented by john santu last updated on 08/Apr/20

super easy sir

$${super}\:{easy}\:{sir} \\ $$$$ \\ $$

Commented by MJS last updated on 08/Apr/20

in this case, it′s easy to see but nearly  impossible  to calculate. you′d have to square 3 times  and then solve an equation of 4^(th)  degree in  x^2 , I′m not sure if I would be able to find the  root x^2 =a^2 +b^2 +c^2  in all that mess...

$$\mathrm{in}\:\mathrm{this}\:\mathrm{case},\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{but}\:\mathrm{nearly}\:\:\mathrm{impossible} \\ $$$$\mathrm{to}\:\mathrm{calculate}.\:\mathrm{you}'\mathrm{d}\:\mathrm{have}\:\mathrm{to}\:\mathrm{square}\:\mathrm{3}\:\mathrm{times} \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{solve}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{degree}\:\mathrm{in} \\ $$$${x}^{\mathrm{2}} ,\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{I}\:\mathrm{would}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{root}\:{x}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:\mathrm{in}\:\mathrm{all}\:\mathrm{that}\:\mathrm{mess}... \\ $$

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